Rectangular Disk with Circular Hole Period Question

In summary, the conversation discusses the problem of finding the period of oscillations for a circular disk with a rectangular hole suspended as a pendulum. The disk has a radius of 0.620 m, mass of 0.470 kg, and moment of inertia of 1.60E-1 kgm2. The center of mass is located at a distance of 0.120 m from the center of the circle. The suggested method is to use the formula T=2∏√I/(mgd) and make an approximation for small angles. After realizing the mistake of using the wrong distance, the correct answer is obtained. The conversation concludes with the advice to always go back to first principles and do a derivation when
  • #1
JohnnyCollins
4
0

Homework Statement


A circular disk with a rectangular hole has a radius of 0.620 m and mass of 0.470 kg. It is suspended by a point on its perimeter as shown in the figure. The moment of inertia about this point is I_p = 1.60E-1 kgm2. Its center of mass is located at a distance of s=0.120 m from the center of the circle as shown. If the disk is allowed to oscillate side to side as a pendulum, what is the period of oscillations


Homework Equations



ω=2∏f
f=1/T
I_g=1/2mr^2

The Attempt at a Solution


I'm lost on this question, any help would be greatly appreciated!
 
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  • #2
Since you are stuck - go back to first principles ... start by drawing a free-body diagram for an arbitrary angular displacement and work out the torques. Fortunately you have been given the moment of inertia and the center of mass. Make an approximation for small angles and solve for T.
 
Last edited:
  • #3
Simon Bridge said:
Since you are stuck - go back to first principles ... start by drawing a free-body diagram for an arbitrary angular displacement and work out the torques. Fortunately you have been given the moment of inertia and the center of mass. Make an approximation for small angles and solve for T.

So i took your advice and used the formula T=2∏√I/(mgd) and used 0.120 m for my distance, but I still ended up with the wrong answer. Am I on the right track ?
 
  • #4
d, in your formula, is the distance from the pivot to the center of mass.
0.120m is the distance from the center of the circle to the center of mass.
spot the difference.
 
  • #5
Simon Bridge said:
d, in your formula, is the distance from the pivot to the center of mass.
0.120m is the distance from the center of the circle to the center of mass.
spot the difference.

Got it, thanks a lot for your help!
 
  • #6
No worries.

When you get stuck - go back to first principles and do a derivation.
It's amazing how naive you can be about this and still get results - just "describe the system in math and then fiddle the math" is very powerful.

In the process you may discover that you could have used a short-cut ... but having gone back over the basics deepens your understanding of the process.
Happy hacking.
 

Related to Rectangular Disk with Circular Hole Period Question

1. What is a "Rectangular Disk with Circular Hole Period Question"?

A "Rectangular Disk with Circular Hole Period Question" is a mathematical problem involving a rectangular disk with a circular hole in the center. The goal is to find the period of the disk's rotation, which is the time it takes for the disk to complete one full revolution.

2. How is the period of the disk's rotation calculated?

The period of the disk's rotation can be calculated using the formula T = 2π√(I/mr^2), where T is the period, I is the moment of inertia, m is the mass of the disk, and r is the radius of the circular hole. This formula is derived from the principles of rotational motion and can be solved using basic calculus.

3. What factors affect the period of the disk's rotation?

The period of the disk's rotation is affected by several factors, including the moment of inertia (which depends on the shape and distribution of mass of the disk), the mass of the disk, and the radius of the circular hole. These factors can be adjusted to change the period of the disk's rotation.

4. What real-world applications does this problem have?

This problem has applications in various fields, such as engineering, physics, and mathematics. For example, it can be used to calculate the period of rotation for objects like flywheels, gyroscopes, and even planets. It also has practical applications in designing and optimizing machinery and structures that involve rotating parts.

5. Are there any practical limitations to this problem?

Like any mathematical problem, there may be some practical limitations to consider when applying this problem in real-world scenarios. For instance, the assumptions and simplifications made in the calculations may not perfectly reflect the actual physical system. Additionally, the accuracy of the calculated period may be affected by factors such as friction, air resistance, and imperfections in the materials used.

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